Question

Помогите, пожалуйста! Решите данные системы неравенств:

Answer 1

$1); ; left { {{x^2<9} atop {(3x-5)^2<1}} right. ; left { {{x^2-9<0} atop {(3x-5)^2-1<0}} right. ; left { {{(x-3)(x+3)<0} atop {(3x-5-1)(3x-5+1)<0}} right. ; left { {{xin (-3,3)} atop {(3x-6)(3x-4)<0}} right. \\left { {{xin (-3,3)} atop {xin (1frac{1}{3};2)}} right. ; ; to ; ; xin (1frac{1}{3}, ;2)$

$2); ; left { {{x^2geq 4} atop {(2-5x)^2geq 16}} right. ; left { {{x^2-4geq 0} atop {(2-5x)^2-16geq 0}} right. ; left { {{(x-2)(x+2)geq 0} atop {(2-5x-4)(2-5x+4)geq 0}} right. \\left { {{xin (-infty ,-2, ]cup [, 2,+infty )} atop {-(2+5x)(6-5x)geq 0}} right. ; left { {{xin (_infty ,-2, ]cup [, 2,+infty )} atop {xin (-infty ,frac{2}{5}, ]cup , [frac{6}{5},+infty )}} right. ; to ; xin (-infty ,frac{2}{5}, ]cup , [1frac{1}{5},+infty )$

$3); ; left { {{x^2leq 16} atop {(x-2)^2geq 9}} right. ; left { {{x^2-16leq 0} atop {(x-2)^2-9geq 0}} right. ; left { {{(x-4)(x+4)leq 0} atop {(x-2-3)(x-2+3)geq 0}} right. ; left { {{(x-4)(x+4)leq 0} atop {(x-5)(x+1)geq 0}} right. \\left { {{xin [-4,4, ]} atop {xin (-infty , 1, ]cup [, 5,+infty )}} right. ; ; to ; ; xin [-4,1, ]$

Answer 2

Спасибо огромное!